Answer: 149,150,151
Step-by-step explanation:
1.st solution
Let the smallest number is x.
So next consecutive number is x+1
So the third consecutive number is x+2
So the sum of these three numbers is:
x+x+1+x+2=450
3x+3=450
3x=447
x=149 - first number
Second number is 149+1=150
Third number is 150+1=151
2-nd solution
The sum of 3 numbers in row is 450. It means that average number is 450:3=150 - is the number which is in the middle
So the first one number is 150-1=149 and the third number is 150+1=151
Edwin has 3 1 2 gallons of green paint. He uses 2 3 of the paint to paint his bedroom. He uses the rest of the paint for a mural in his garage. How much paint does Edwin use for the mural?
Answer:
1 1/6
Step-by-step explanation:
Edwin used a quantity of 2.833 gallons of paint for the mural which is the difference between the quantity of paint at the beginning and the used for the bedroom.
We have been given that Edwin has 3 1/2 gallons of green paint. He uses 2/3 of the paint to paint his bedroom. He uses the rest of the paint for a mural in his garage.
What is the fraction?A fraction is defined as a numerical representation of a part of a whole that represents a rational number.
To determine the quantity of paint Edwin used for the mural.
The quantity of paint Edwin used for the mural is the difference between the quantity of paint at the beginning and the used for the bedroom.
The quantity of paint used for the mural = 3 1/2 gallons - 2/3 gallons
The quantity of paint used for the mural = 3.5 - 0.66
The quantity of paint used for the mural = 2.833
Thus, Edwin used a quantity of 2.833 gallons of paint for the mural.
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how to solve the birth rate in a certain country in 1994 was 14.6 births per thousand population. in 2004 the birth rate was 14.32 births per thousand. let x represent years after 1994 and y represent the birth rate. assume the relationship is linear
Answer:
[tex]y(x)=-0.028x+14.6[/tex]
Step-by-step explanation:
We are to write a linear equation that relates y in terms of x
The Birth Rate in 1994 = 14.6 births per thousand population.
The Birth Rate in 2004 = 14.32 births per thousand population.
A linear equation is of the form y=mx+b, where:
x=Number of Years after 1994y=the birth ratem=Birth rate per yearStep 1: Determine the birth rate per year
In 1994, x=0, y=14.6 thousands
In 2004, x=10, y=14.32 thousands
[tex]m=\dfrac{14.32-14.6}{10-0}\\=\dfrac{-0.28}{10}\\m=-0.028[/tex]
Substituting m into our linear equation, we have:
[tex]y(x)=-0.028x+b[/tex]
When x=10, y=14.32
[tex]14.32=-0.028(10)+b\\b=14.32+0.28\\b=14.6[/tex]
Therefore, a linear equation that relates y in terms of x is:
[tex]y(x)=-0.028x+14.6[/tex]
hey guys, can you help me please
=========================================================
Work Shown:
The green triangle in the back has height 2.6 and an unknown base x. Half of this is x/2, which I'll call y. So y = x/2.
The green triangle in the back is split along the vertical dotted line to get two right triangles. The base of each right triangle is y = x/2.
Use the Pythagorean theorem to find y. Use that to find x
a^2+b^2 = c^2
y^2+(2.6)^2 = (3.2)^2
y^2 + 6.76 = 10.24
y^2 = 10.24 - 6.76
y^2 = 3.48
y = sqrt(3.48) .... apply square root
y = 1.8654758 approximately
x/2 = 1.8654758
x = 2*1.8654758
x = 3.7309516 also approximate
The base of the triangle is roughly 3.7309516 meters
We can now find the area of one green triangle
area of triangle = base*height/2 = 3.7309516*2.6/2 = 4.85023708
two triangles have approximate area 2*(4.85023708) = 9.70047416
----------------------------------
So far we've only considered the triangular faces. There are 3 more faces which are the rectangular sides. These are known as the lateral sides.
One way to get the lateral surface area is to multiply the perimeter of the triangle by the depth of the prism
perimeter of triangle = (side1)+(side2)+(side3)
perimeter = 3.7309516 + 3.2 + 3.2
perimeter = 10.1309516
lateral surface area = (depth)*(perimeter)
lateral surface area = (8.26)*(10.1309516)
lateral surface area = 83.681660216
----------------------------------
The last step is to add this lateral surface area onto the area of the two triangles to get the full surface area
surface area = (triangular area) + (lateral surface area)
surface area = (9.70047416) + (83.681660216)
surface area = 93.382134376
surface area = 93.382 square cm
Which leader was a member of the Kikuyu tribe?
A. Kwame Nkrumah
B. Marcus Garvey
C. Mohandas Gandhi
D. Jomo Kenyatta
Answer:
Jomo Kenyatta
Step-by-step explanation:
Jomo Kenyatta was a Kenyan politician, who was one of the first African anti-colonial figures. He became the prime minister of Kenya from 1963 to 1964, and after Kenyan independence in 1964, he became president of Kenya. Jomo Kenyatta was born into a family of Kikuyu farmers in Kiambu, present day Kenya which was then, British East Africa. He had his basic schooling in a missionary school before proceeding to study at Moscow's Communist University of the Toilers of the East, University College London, and the London School of Economics.
Answer:
Jomo Kenyatta
Step-by-step explanation:
took the test
A cylinder is inscribed in a right circular cone of height 4.5 and radius (at the base) equal to 5.5 . What are the dimensions of such a cylinder which has maximum volume? Its radius is equation editorEquation Editor and its height is
Answer:
r = 3.667
h = 1.5
Step-by-step explanation:
Given:-
- The base radius of the right circular cone, R = 5.5
- The height of the right circular cone, H = 4.5
Solution:-
- We will first define two variables that identifies the volume of a cylinder as follows:
r: The radius of the cylinder
h: The height of cylinder
- Now we will write out the volume of the cylinder ( V ) as follows:
[tex]V = \pi*r^2h[/tex]
- We see that the volume of the cylinder ( V ) is a function of two variables ( don't know yet ) - ( r,h ). This is called a multi-variable function. However, some multi-variable functions can be reduced to explicit function of single variable.
- To convert a multi-variable function into a single variable function we need a relationship between the two variables ( r and h ).
- Inscribing, a cylinder in the right circular cone. We will denote 5 points.
Point A: The top vertex of the cone
Point B: The right end of the circular base ( projected triangle )
Point C: The center of both cylinder and base of cone.
Point D: The top-right intersection point of cone and cylinder
Point E: Denote the height of the cylinder on the axis of symmetry of both cylinder and cone.
- Now, we will look at a large triangle ( ABC ) and smaller triangle ( ADE ). We see that these two triangles are "similar". Therefore, we can apply the properties of similar triangles as follows:
[tex]\frac{AC}{AE} = \frac{BC}{DE} \\\\\frac{H}{H-h} = \frac{R}{r}[/tex]
- Now we can choose either variable variable to be expressed in terms of the other one. We will express the height of cylinder ( h ) in term of radius of cylinder ( r ) as follows:
[tex]H- h = r\frac{H}{R} \\\\h = \frac{H}{R}*(R-r)[/tex]
- We will use the above derived relationship and substitute into the formula given above:
[tex]V = \pi r^2 [ \frac{H}{R}*(R - r )]\\\\V = \frac{\pi H}{R}.r^2.(R-r)[/tex]
- Now our function of volume ( V ) is a single variable function. To maximize the volume of the cylinder we need to determine the critical points of the function as follows:
[tex]\frac{dV}{dr} = \frac{\pi H}{R}*(2rR-2r^2 - r^2 )\\\\\frac{dV}{dr} = \frac{\pi H}{R}*(2rR-3r^2 ) = 0\\\\(2rR-3r^2 ) = 0\\\\2R -3r = 0\\\\r = \frac{2}{3}*R[/tex]
- We found the limiting value of the function. The cylinder volume maximizes when the radius ( r ) is two-thirds of the radius of the right circular cone.
- We can use the relationship between the ( r ) and ( h ) to determine the limiting value of height of cylinder as follows:
[tex]h = \frac{H}{R} * ( R - \frac{2}{3}R)\\\\h = \frac{H}{3}[/tex]
- The dimension of the inscribed cylinder with maximum volume are as follows:
[tex]r = \frac{2}{3}*5.5 = 3.667\\\\h = \frac{4.5}{3} = 1.5[/tex]
Note: When we solved for the critical value of radius ( r ). We actually had two values: r = 0 , r = 2R/3. Where, r = 0 minimizes the volume and r = 2R/3 maximizes. Since the function is straightforward, we will not test for the nature of critical point ( second derivative test ).
A circular swimming pool has a diameter of 20 ft, the sides are 6 ft high, and the depth of the water is 5 ft. How much work (in ft-lb) is required to pump all of the water out over the side
Answer:
19467649.76 lb-ft^2/s^2
Step-by-step explanation:
diameter of the pool d = 20 ft
radius = d/2 = 20/2 = 10 ft
height of pool side h = 6 ft
depth of water d = 5 ft
the force on the bottom of the pool due to the water in the pool is
F = pgdA
where p = density of water = 62.4 lb/ft^3
g = acceleration due to gravity = 32.17 ft/s^2
Area A = [tex]\pi r^{2}[/tex] = [tex]3.142 * 10^{2}[/tex] = 314.2 ft^2
Force on pool bottom = 64.2 x 32.17 x 5 x 314.2 = 3244608.29 lb-ft/s^2
work done = force times the height the water will be pumped
work = F x h = 3244608.29 x 6 = 19467649.76 lb-ft^2/s^2
The work (in ft-lb) is required to pump all of the water out over the side is :
Given :Diameter of the pool d = 20 ftRadius = d/2 = 20/2 = 10 ftHeight of pool side h = 6 ftDepth of water d = 5 ftFormula:
F = pgdA
p = density of water = 62.4 lb/ft^3
g = acceleration due to gravity = 32.17 ft/s^2
Area A = [tex]\pi r2\\[/tex] = 314.2 ft^2
Force on pool bottom = 64.2 x 32.17 x 5 x 314.2 = 3244608.29 lb-ft/s^2work done = force times the height the water will be pumpedwork = F x h = 3244608.29 x 6 = 19467649.76 lb-ft^2/s^2The work (in ft-lb) is required to pump all of the water out over the side is 19467649.76 lb-ft^2/s^2.
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The management of a chain of frozen yogurt stores believes that t days after the end of an advertising campaign, the rate at which the volume V (in dollars) of sales is changing is approximated by V ' ( t ) = − 26400 e − 0.49 t . On the day the advertising campaign ends ( t = 0 ), the sales volume is $ 170 , 000 . Find both V ' ( 6 ) and its integral V ( 6 ) . Round your answers to the nearest cent.
Answer:
Step-by-step explanation:
Given the rate at which the volume V (in dollars) of sales is changing is approximated by the equation
V ' ( t ) = − 26400 e^− 0.49 t .
t = time (in days)
.v'(6) can be derived by simply substituting t = 6 into the modelled equation as shown:
V'(6) = − 26400 e− 0.49 (6)
V'(6) = -26400e-2.94
V'(6) = -26400×-0.2217
V'(6) = $5852.88
V'(6) = $5,853 to nearest dollars
V'(6) = 585300cents to nearest cent
To get v(6), we need to get v(t) first by integrating the given function as shown:
V(t) = ∫−26400 e− 0.49 t dt
V(t) = -26,400e-0.49t/-0.49
V(t) = 53,877.55e-0.49t + C.
When t = 0, V(t) = $170,000
170,000 = 53,877.55e-0 +C
170000 = 53,877.55(2.7183)+C
170,000 = 146,454.37+C
C = 170,000-146,454.37
C = 23545.64
V(6) = 53,877.55e-0.49(6)+ 23545.64
V(6) = -11,945.63+23545.64
V(6) = $11,600 (to the nearest dollars)
Since $1 = 100cents
$11,600 = 1,160,000cents
The manager of a warehouse would like to know how many errors are made when a product’s serial number is read by a bar-code reader. Six samples are collected of the number of scanning errors: 36, 14, 21, 39, 11, and 2 errors, per 1,000 scans each.
Just to be sure, the manager has six more samples taken:
33, 45, 34, 17, 1, and 29 errors, per 1,000 scans each
How do the mean and standard deviation change, based on all 12 samples?
Answer:
The mean and standard deviation changed to 23.5 and 14.62 respectively, based on all 12 samples.
Step-by-step explanation:
We are given that the Six samples are collected of the number of scanning errors: 36, 14, 21, 39, 11, and 2 errors, per 1,000 scans each.
Representing the data in tabular form;
X [tex]X - \bar X[/tex] [tex](X - \bar X)^{2}[/tex]
36 36 - 20.5 = 15.5 240.25
14 14 - 20.5 = -6.5 42.25
21 21 - 20.5 = 0.5 0.25
39 39 - 20.5 = 18.5 342.25
11 11 - 20.5 = -9.5 90.25
2 2 - 20.5 = -18.5 342.25
Total 1057.5
Now, the mean of these value is given by;
Mean, [tex]\bar X[/tex] = [tex]\frac{\sum X}{n}[/tex]
= [tex]\frac{36+14+21+39+11+2}{6}[/tex]
= [tex]\frac{123}{6}[/tex] = 20.5
Standard deviation formula for discrete distribution is given by;
Standard deviation, [tex]\sigma[/tex] = [tex]\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }[/tex]
= [tex]\sqrt{\frac{1057.5 }{6-1} }[/tex] = 14.54
Now, the manager has six more samples taken:
33, 45, 34, 17, 1, and 29 errors, per 1,000 scans each
So, the modified table would be;
X [tex]X - \bar X[/tex] [tex](X - \bar X)^{2}[/tex]
36 36 - 23.5 = 12.5 156.25
14 14 - 23.5 = -9.5 90.25
21 21 - 23.5 = -2.5 6.25
39 39 - 23.5 = 15.5 240.25
11 11 - 23.5 = -12.5 156.25
2 2 - 23.5 = -21.5 462.25
33 33 - 23.5 = 9.5 90.25
45 45 - 23.5 = 21.5 462.25
34 34 - 23.5 = 10.5 110.25
17 17 - 23.5 = -6.5 42.25
1 1 - 23.5 = -22.5 506.25
29 29 - 23.5 = 5.5 30.25
Total 2353
Now, the mean of these value is given by;
Mean, [tex]\bar X[/tex] = [tex]\frac{\sum X}{n}[/tex]
= [tex]\frac{36+14+21+39+11+2+33+45+34+17+1+29}{12}[/tex]
= [tex]\frac{282}{12}[/tex] = 23.5
Standard deviation formula for discrete distribution is given by;
Standard deviation, [tex]\sigma[/tex] = [tex]\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }[/tex]
= [tex]\sqrt{\frac{2353 }{12-1} }[/tex] = 14.62
How many pairs are shown ?????????
Answer:8 i ithink
Step-by-step explanation:
Answer:
12, go for 24.
Step-by-step explanation:
There are 6 sides of a cube.
There are 2 pairs of parallel line segments for each side.
6 x 2 = 12
Although that answer is not there, you should go for 24. Since there are 2 variables for each line segment, 12 x 2 = 24. Not sure, hope this helps.
:/
An article discussed how anger intensity in children's tantrums could be related to tantrum duration as well as behavioral indicators such as shouting, stamping, and pushing or pulling. The following frequency distribution was given:0-<2: 136 2- <4: 92 4-<11: 7211-<20 26 20-<30 7 30-<40: 3Required:Draw the histogram and then comment on any interesting features.
Answer:
Increase in age is indirectly proportional to anger intensity in children's tantrums
Step-by-step explanation:
Let's begin by reproducing the given information into a frequency distribution table (shown below):
Age Group Frequency
0 - <2 136
2 - <4 92
4 - <11 72
11 - <20 26
20 - <30 7
30 - <40 3
Observing the dataset above, we will see a trend; increase in age is inversely proportional to anger intensity. Kindly note that the histogram is attached as a picture.
Analysis of Histogram
Age group 1 (infants) have the highest indices of anger intensity in tantrums. This is expected because they do not know much yet and have an entitlement mentality.
Age group 2 (infants) also have a pretty high amount of anger intensity in children tantrums. They have transitioned from infancy but are yet immature & can easily flare up when disgruntled.
Age group 3 (children) are not far off from the previous age bracket, also having a significantly fairly high anger intensity. They are advancing in years & as they do, their emotional rage to tantrums is reducing.
Age group 4 (pre-teens/teens) experience a significant and drastic drop in emotional rage due to children's tantrums. They are maturing and are learning to communicate in less fitful ways. They now know better than to express their displeasure like infants.
Age group 5 (youths) also see a drastic drop in such anger intensity. The reason is not far-fetched; they are more mature and can express their displeasure in more mature ways.
Age group 6 (Adults) almost have almost a zero or non-existent such attitude among them. At this point, they have come to understand & know that they can get much more done with words than such outlandish methods.
In conclusion, as a child advances in years (changing from one age group to another → from infant to adult), their anger intensity takes a nosedive. We rightly interpret the histogram when we say that physical maturity is inversely proportional to anger intensity in children's tantrums.
⅝ of a school's population are girls. There are 129 boys. If each classroom can hold 25 students. How many classroom does the school have ?
Answer:
AT least 14 classrooms to hold the total number of students
Step-by-step explanation:
Since we don't know the numer of girls in the school, let's call it "x".
What we know is that the number of girls plus the number of boys gives the total number of students. This is:
x + 129 = Total number of students
Now, since 5/8 of the total number of students are girls, and understanding that 5/8 = 0.625 in decimal form, then we write the equation that states:
"5/8 of the school's population are girls" as:
0.625 (x + 129) = x
then we solve for "x":
0.625 x + 0.625 * 129 = x
0.625 * 129 = x - 0.625 x
80.625 = x (1 - 0.625)
80.625 = 0.375 x
x = 80.625/0.375
x = 215
So now we know that the total number of students is: 215 + 129 = 344
and if each classroom can hold 25 students, the number of classroom needed for 344 students is:
344/25 = 13.76
so at least 14 classrooms to hold all those students
Tyler drew a figure that has two pairs of equal sides, four angles formed by perpendicular lines, and two pairs of parallel sides. What geometric term best describes the figure Tyler drew? What geometric term best describes the figure Tyler drew?
Answer:
A shape with two pairs of parallel lines, perpendicular lines, and two pairs of equal sides can be best described as a rectangle.
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95% confidence interval for the population mean amount spent on groceries (in dollars) by Connecticut families. Assume the data come from a normal distribution
210 23 350 112 27 175 275 50 95 450
Answer:
The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).
Step-by-step explanation:
The data for the amount of money spent weekly on groceries is as follows:
S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}
n = 10
Compute the sample mean and sample standard deviation:
[tex]\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7[/tex]
[tex]s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702[/tex]
It is assumed that the data come from a normal distribution.
Since the population standard deviation is not known, use a t confidence interval.
The critical value of t for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:
[tex]t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262[/tex]
*Use a t-table.
Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}[/tex]
[tex]=176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)[/tex]
Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).
What type of error is present in the underlined
sentence?
Which is the best revision to fix the error?
Answer:
Type of error: Run-on(comma splice).
Best revision to fix it: Adding a semicolon after beginners .
Explanation:
A run-on sentence is described as a sentence in which two independent clauses are joined inappropriately. It could be either comma splice where the two independent clauses are incorrectly linked using a comma or fused sentence when the two clauses run-on without employing appropriate coordinating conjunction or punctuation marks to separate the two ideas.
In the given sentence, it exemplifies a comma splice type of run-on sentence error. To fix this error, a semicolon after 'beginners' can be employed instead of a comma. This will help in connecting the two ideas appropriately where the first idea leads the second. Thus, the final sentence reads as:
'The guitar is another excellent instrument for beginners; however, it takes more practice than a recorder.'
Answer:
Many people play a musical instrument music can be soothing. A lot of schools teach the recorder; it is inexpensive and easy to play. The guitar is another excellent instrument for beginners, it takes more practice than a recorder.
What type of error is present in the underlined sentence?
✔ run-on
Which is the best revision to fix the error?
✔ adding a semicolon after instrument
Step-by-step explanation:
Adult male heights are a normal random variable with mean 69 inches and a standard deviation of 3 inches. Find the height, to the nearest inch, for which only 8 percent of adult males are taller (i. find the 92nd percentile)
Answer:
The height (corresponding to the [tex] \\ 92^{nd}[/tex] percentile) is (to the nearest inch) 73 inches (and, approximately, only 8% of adult males are taller than this height.)
Step-by-step explanation:
Roughly speaking, the [tex] \\ 92^{nd}[/tex] percentile is the x value (in the distribution) for which 92% of the observations in the [normal] distribution are below this x value, and 8% of the observations are above this x value.
To answer this question, we already know that:
Heights are a normal random variable, i.e, it follows a normal distribution.The mean for this distribution is [tex] \\ \mu = 69[/tex] inches.The standard deviation is [tex] \\ \sigma = 3[/tex] inches.Strategy for solving the question
For solving this, we have to use here the following key concepts: z-scores, the cumulative standard normal distribution, and the cumulative standard normal table.
Z-scores
To find the [tex] \\ 92^{nd}[/tex] percentile, we can use z-scores or standardized values. A z-score is a value that tells us the distance in standard deviations units from the mean. When the z-score is positive, it means that the value is above the mean. A negative indicates that the z-score is below the mean. The formula to obtain a z-score is as follows:
[tex] \\ z = \frac{x - \mu}{\sigma}[/tex] [1]
Where
z is the z-score.x is the raw score.[tex] \\ \mu[/tex] is the mean.[tex] \\ \sigma[/tex] is the standard deviation.Cumulative standard normal distribution and corresponding table
We still need to know the corresponding z-score, z, for the cumulative probability of 92%. For this, we have to consult the standard normal table, available on the Internet or in any Statistics books.
In this case, we look in the different columns of the standard normal table a probability value (exact or approximate) to 0.92 and then find the value for z that corresponds to this probability. The value for z is between 1.40 (0.91924) and 1.41 (0.92073).
Using z = 1.40 in [1], we have:
[tex] \\ z = \frac{x - \mu}{\sigma}[/tex]
[tex] \\ 1.40 = \frac{x - 69}{3}[/tex]
Then, solving for x:
Multiplying by 3 at each side of the equation:
[tex] \\ 1.40 * 3 = x - 69[/tex]
Adding 69 at both sides of the equation:
[tex] \\ (1.40 * 3) + 69 = x[/tex]
[tex] \\ x = (1.40 * 3) + 69[/tex]
[tex] \\ x = 4.20 + 69[/tex]
[tex] \\ x = 73.20[/tex]
That is, the [tex] \\ 92^{nd}[/tex] percentile is 73.20 inches, and to the nearest inch, this percentile is 73 inches.
This result indicates that, approximately, 92% of the heights are below 73 inches, and only 8% of heights are taller than this height.
The shaded area in the graph below shows an area of 0.08076 (8.076%) for 73.20 inches.
A large toiletry distributor claims that 35% of all individuals who purchase toilet paper from the stores that carry its product choose original toilet paper, 28% choose sensitive toilet paper, 20% choose ultra-strong toilet paper, and 17% choose ultra-soft toilet paper. To investigate this claim, researchers collected data from a random sample of customers in a large city. The results were 170 packages of original, 105 sensitive, 80 ultra-strong, and 45 ultra-soft toilet paper purchases. Are the data from the sample consistent with the distributor's claim
p-value < 0.05, reject null hypothesis and we can conclude that data is not consistent with distributor's claim.
Given, data of toiletry distributors.
Total number of observations
= 170 + 105 + 80 + 45
= 400
Expected count = [tex]p_i \times 400[/tex]
Calculation table is attached below.
Test statistic:
Chi square score
= 6.429 + 0.438 + 0.000 + 7.779
= 14.645
Degree of freedom:
df = 4-1
=3
p-value = CHIDIST(14.645,3)
= 0.00215
Therefore , p-value < 0.05
Reject null hypothesis and we can conclude that data is not consistent with distributor's claim.
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Approximately 10% of all people are left-handed. If 200 people are randomly selected, what is the expected number of left-handed people? Round to the whole number. Do not use decimals. Answer:
Answer:
N(L) = 20
The expected number of left handed people is 20.
Step-by-step explanation:
Given;
Percentage of left handed people P(L) = 10%
Total number of selected people N(T) = 200
The Expected number of left handed people N(L) is;
N(L) = Total number of selected people × Percentage of left handed people/100%
N(L) = N(T) × P(L)/100%
Substituting the given values;
N(L) = 200 × 10%/100%
N(L) = 200 × 0.1
N(L) = 20
The expected number of left handed people is 20.
Suppose that there are six universities and each will produce five mathematics Ph.D.s this year, and there are five colleges that will be hiring seven, seven, six, six, five math Ph.D.s, respectively. No college will hire more than one Ph.D. from any given university. Will all the Ph.D.s get a job? Explain.
Answer:
No. If a collegue hire 7 math Ph.D.s, at least 2 will be from the same university.
The only way to have a job for all of them is that all colleges hire 6 math Ph.D.s, one from each university.
Step-by-step explanation:
If the condition is that no college will hire more than one Ph.D. from any given university, no college can hire more than 6 math Ph.D.s, as there are only 6 universities to choose from. If they hire seven, they have to repeat some university, violating the condition.
Then, the only way to have a job for all of them is that all colleges hire 6 math Ph.D.s, one from each university.
Which statements are true of the function f(x) = 3(2.5)x? Check all that apply
Answer:
The function is exponential.
The function increases by a factor of 2.5 for each unit increase in x.
The domain of the function is all real numbers
The true statements are:
[tex]\mathbf{y = 3(2.5)^x}[/tex] is an exponential functionThe function represents an exponential growthThe domain of the function is the set of all real numbersThe function is given as:
[tex]\mathbf{y = 3(2.5)^x}[/tex]
An exponential function is represented as:
[tex]\mathbf{y = ab^x}[/tex]
Where: a represents the initial value, and b represents the rate
This means that:
[tex]\mathbf{y = 3(2.5)^x}[/tex] is an exponential function
By comparison:
[tex]\mathbf{b = 2.5}[/tex]
When b > 0, then the function represents an exponential growth
2.5 is greater than 0.
So, the function represents an exponential growth
Lastly, there is no restriction to the values of x.
So, the domain of the function is the set of all real numbers
Read more about exponential functions at:
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A utility company offers a lifeline rate to any household whose electricity usage falls below 240 kWh during a particular month. Let A denote the event that a randomly selected household in a certain community does not exceed the lifeline usage during January, and let B be the analogous event for the month of July (A and B refer to the same household). Suppose P(A) = 0.7, P(B) = 0.5, and P(A ∪ B) = 0.8. Compute the following.a. P(A ∩ B).b. The probability that the lifeline usage amount is exceeded in exactly one of the two months.
Complete question is;
A utility company offers a lifeline rate to any household whose electricity usage falls below 240 kWh during a particular month. Let A denote the event that a randomly selected household in a certain community does not exceed the lifeline usage during January, and let B be the analogous event for the month of July (A and B refer to the same household). Suppose P(A) = 0.7, P(B) = 0.5, and P(A ∪ B) = 0.8. Compute the following.
a. P(A ∩ B).
b. The probability that the lifeline usage amount is exceeded in exactly one of the two months.
Answer:
A) 0.4
B) 0.4
Step-by-step explanation:
We are given;
P(A) = 0.7, P(B) = 0.5, and P(A ∪ B) = 0.8
A) To solve this question, we will use the the general probability addition rule for the union of two events which is;
P(A∪B) = P(A) + P(B) − P(A∩B)
Making P(A∩B) the subject of the equation, we have;
P(A∩B) = P(A) + P(B) − P(A∪B)
Thus, plugging in the relevant values, we have;
P(A∩B) = 0.7 + 0.5 - 0.8
P(A∩B) = 0.4
B)The probability that the lifeline usage amount is exceeded in exactly one of the two months can be described in terms of A and B as:
P(A but not B) + P(B but not A) = P(A∩B') + P(B∩A')
where;
A' is compliment of set A
B' is compliment of set B
Now,
P(A∩B') = 0.7 − 0.4 = 0.3
P(B∩A') = 0.5 − 0.4 = 0.1
Thus;
P(A but not B) + P(B but not A) = 0.1 + 0.3 = 0.4
∠A and ∠B are supplementary, and ∠A and ∠C are supplementary. Which conclusion is valid? Select one: A. ∠B and ∠C are supplementary. B. ∠B and ∠C are acute. C. ∠B and ∠C are complementary. D. ∠B and ∠C are congruent.
Option D is the correct answer.
Answer:
D. ∠B and ∠C are congruent.
Step-by-step explanation:
Since, ∠A and ∠B are supplementary.
Therefore,
∠A + ∠B = 180°.....(1)
Since, ∠A and ∠C are supplementary.
Therefore,
∠A + ∠C = 180°.....(2)
From equations (1) & (2)
∠A + ∠B = ∠A + ∠C
=> ∠B = ∠C
Hence, ∠B and ∠C are congruent.
) Let an denote number of n-digit ternary sequences (sequences of 0,1 and 2) which have no consecutive 0’s in them. Find a recurrence relation for an. (Do not solve the recurrence. However, depending on the order of the recurrence, provide a sufficient number of initial conditions. )
Answer:
The recurrence relation for aₙ is aₙ = 2aₙ - 1 + 2aₙ -2 ; is n≥ 3 with the initial conditions as a₁ =3; a₂ = 8
Step-by-step explanation:
Solution
Recurrence relation for n - digit ternary sequence with no occurrence of consecutive 0's in them.
Ternary sequence is sequence with each of digits either 0, 1 or 2.
Now
Let aₙ = denote the number of n - digit ternary sequence with no occurrence of consecutive 0's in them.
Let us first find few initial values of aₙ
For n = 1
a₁ represent the number of 1- digit ternary sequence with no occurrence of consecutive 0's in them.
This 1-digit sequence can be either 0 or 1 or 2.
Thus,
a₁ = 3
For n =2
a₂ represent the number of 2- digit ternary sequence with no occurrence of consecutive 0's in them.
This 2-digit sequence can have either 0 or 1 or 2 as each of its two digit, but making sure that there are no two consecutive 0 in the sequence.
here are " 9 " 2-digit ternary sequence ........... (three choices for 1st digit and three choices for 2nd digit)
But one of these 9 sequence there are consecutive 0's .... (00)
So we eliminate this one sequence.
So, a₂ = 8
Now
let us find the recurrence relation
Fir n ≥ 3
aₙ s the number of n - digit ternary sequence with no occurrence of the consecutive 0's in them.
For the first case: if 1st digit of this n - digit ternary sequence is 1 or 2
Let assume the 1st digit of this n - digit ternary sequence is 1.
Then for remaining n - 1 digits of this n - digit ternary sequence we have to make sure that there is no consecutive 0's in them.
For example, we have to form a n-1-digit ternary sequence with no occurrence of consecutive 0's in them which is by definition aₙ -1
So,
The number of n - digit ternary sequence with no occurrence of consecutive 0's in them if 1st digit of this sequence is 1 is aₙ -1.
Likewise, the number of n - digit ternary sequence with no occurrence of consecutive 0's in them if 1st digit of this sequence is 2 is aₙ -1.
So
If 1st digit of this n - digit ternary sequence is 1 or 2, then the number of n - digit ternary sequence with no occurrence of consecutive 0's in them is shown as:
aₙ-1 + aₙ -1 = 2aₙ -1
For the second case: if 1st digit of this n - digit ternary sequence is 0
If 1st digit of this n - digit ternary sequence is 0, then the next digit cannot be 0 as well because that would make two consecutive 0's in the sequence Thus,
If 1st digit of this n - digit ternary sequence is 0, the next term can be either 1 or 2.
So there are 2 choices for 2nd digit.
After this there are more n-2 digits.
Then
For remaining n - 2 digits of this n - digit ternary sequence we have to make sure that there is no consecutive 0's in them
For example, we have to form a n-2-digit ternary sequence with no occurrence of consecutive 0's in them. which is by definition aₙ - 2.
Now,
The total number of sequence in this case is given as:
2aₙ -2........... (2 choices for 2nd digit and aₙ - 2 choices for remaining n-2 digit)
Hence
The number of n - digit ternary sequence with no occurrence of consecutive 0's in them if 1st digit of this sequence is 0 is aₙ = 2aₙ - 1 + 2aₙ -2 which is n≥ 3
Now,
The recurrence relation for aₙ is shown below:
aₙ = 2aₙ - 1 + 2aₙ -2; is n≥ 3
With the initial conditions as a₁ =3; a₂ = 8
Let U be the 3 2 matrix [0.45 0.42, 0.25 0.35, 0.15 0.15]. The first column of U lists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for manufacturing product C. The first row is the cost of materials, the second row is the cost of labor, and the third row is the cost of overhead. Let q1 be a vector in set of real numbers R2 that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let q2, q3 , and q4 be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic desciption of the data in the matrix UQ, where Upper Q = [q1 q2 q3 q4].A. The 4 columns of UQ list the profit made from selling products B and C during the 4 quarters of the year. B. The 3 rows of UQ list the costs for materials, labor, and overhead used to manufacture products B and C for the year. C. The 4 columns of UQ list the total costs for materials, labor, and overhead used to manufacture products B and C during the 4 quarters of the year. D. The 4 columns of UQ list the total number of each product manufactured during the 4 quarters of the year.
Answer:
C. The 4 columns of UQ list the total costs for materials, labor, and overhead used to manufacture products B and C during the 4 quarters of the year.
Step-by-step explanation:
[tex]U=\left\begin{array}{ccc}\\$Cost of Materials\\ $Cost of Labour\\ $Cost of Overhead\end{array}\right\left(\begin{array}{cc}$Product B&$Product C\\0.45&0.42\\ 0.25&0.35\\ 0.15&0.15\end{array}\right)[/tex]
[tex]q_1[/tex] is a vector in the set of real numbers [tex]R^2[/tex] that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year.
Therefore:
[tex]UQ=\left\begin{array}{ccc}\\$Cost of Materials\\ $Cost of Labour\\ $Cost of Overhead\end{array}\right\left(\begin{array}{cc}$Product B&$Product C\\0.45&0.42\\ 0.25&0.35\\ 0.15&0.15\end{array}\right)\left(\begin{array}{ccc}q_{1B}\\q_{1C}\end{array}\right)\left(\begin{array}{ccc}q_{2B}\\q_{2C}\end{array}\right)\left(\begin{array}{ccc}q_{3B}\\q_{3C}\end{array}\right)\left(\begin{array}{ccc}q_{4B}\\q_{4C}\end{array}\right)[/tex]
[tex]=\left(\begin{array}{c|c|c|c}q_1&q_2&q_3&q_4\\0.45q_{1B}+0.42q_{1C}&0.45q_{2B}+0.42q_{2C}&0.45q_{3B}+0.42q_{3C}&0.45q_{4B}+0.42q_{4C}\\0.25q_{1B}+0.35q_{1C}&0.25q_{2B}+0.35q_{2C}&0.25q_{3B}+0.35q_{3C}&0.25q_{4B}+0.35q_{4C}\\0.15q_{1B}+0.15q_{1C}&0.15q_{2B}+0.15q_{2C}&0.15q_{3B}+0.15q_{3C}&0.15q_{4B}+0.15q_{4C}\end{array}\right)[/tex]Therefore, UQ has 4 columns and 3 rows.
The 4 columns of UQ list the total costs for materials(Row 1), labor(Row 2), and overhead(Row 3) used to manufacture products B and C during the 4 quarters of the year.
What number should go in the space? Multiplying by 0.65 is the same as decreasing by _____%
Answer: 35%
Step-by-step explanation:
If no is 10, 10 x 0.65 = 6.5. OR
10 - 35% of 10 = 6.5
Multiplying by 0.65 is the same as decreasing by 35%
Conversion of statements into algebraic expression:To convert the statement into algebraic expression choose the variables first.Then form the expression or equation as per given statements.
Let the number is 'a' and percentage decrease is 'b',
Expression for the given statement will be,
a × 0.65 = a - (b% of a)
[tex]0.65a=a(1-\frac{b}{100})[/tex]
[tex]0.65=1-\frac{b}{100}[/tex]
[tex]\frac{b}{100}=1-0.65[/tex]
[tex]b=100(0.35)[/tex]
[tex]b=35[/tex]
Therefore, Multiplying by 0.65 is the same as decreasing by 35%.
Learn more about the Algebraic expressions for the statements here,
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Consider the differential equation4y'' − 4y' + y = 0; ex/2, xex/2.Verify that the functions ex/2 and xex/2 form a fundamental set of solutions of the differential equation on the interval (−[infinity], [infinity]).The functions satisfy the differential equation and are linearly independent since W(ex/2, xex/2) =
Step-by-step explanation:
Let y1 and y2 be (e^x)/2, and (xe^x)/2 respectively.
The Wronskian of them functions be
W = (y1y2' - y1'y2)
y1 = (e^x)/2 = y1'
y2 = (xe^x)/2
y2' = (1/2)(x + 1)e^x
W = (1/4)(x + 1)e^(2x) - (1/4)xe^(2x)
= (1/4)(x + 1 - x)e^(2x)
W = (1/4)e^(2x)
Since the Wronskian ≠ 0, we conclude that functions are linearly independent, and hence, form a set of fundamental solutions.
Answer:
W = (1/4)(x + 1)e^(2x) - (1/4)xe^(2x)
= (1/4)(x + 1 - x)e^(2x)
W = (1/4)e^(2x)
Step-by-step explanation:
Suppose you start at the origin, move along the x-axis a distance of 7 units in the positive direction, and then move downward along the z-axis a distance of 9 units. What are the coordinates of your position
Answer:
The coordinates of the new position are
(x, y, z) = (7, 0, -9)
Step-by-step explanation:
So assuming a 3D plane x, y, z
The starting point is the origin that is
(x, y, z) = (0, 0, 0)
Move a distance of 7 units along the positive x-axis
(x, y, z) = (7, 0, 0)
Then move a distance of 9 units downwards along the z-axis
downward = negative and upward = positive
(x, y, z) = (7, 0, -9)
Therefore, the coordinates of the new position are
(x, y, z) = (7, 0, -9)
Refer to the attached plot where the above coordinates are plotted on a 3D surface using an online 3D plotter.
Each small box corresponds to 1 unit.
AC =
Round your answer to the nearest hundredth.
A
5
35
B
C
Answer:
2.87 = AC
Step-by-step explanation:
Since this is a right triangle we can use trig functions
sin theta = opp / hyp
sin 35 = AC /5
5 sin 35 = AC
2.867882182= AC
To the nearest hundredth
2.87 = AC
Five pulse rates are randomly selected from a set of measurements. The five pulse rates have a mean of 65.4 beats per minute. Four of the pulse rates are 57, 55, 62, and 83. a. Find the missing value. b. Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected. How many of the n values can be freely assigned before the remaining values are determined? (The result is referred to as the number of degrees of freedom.)
Answer:
a) 70b) n-1Step-by-step explanation:
Mean is defined as the sum total of data divided by the total number of data.
[tex]\overline x = \frac{\sum Xi}{N}[/tex]
Xi are individual data
N is the total number of data = 5
Given the mean of the pulse rate = 65.4
a) Let the 5pulse rates be our data as shown: 57, 55, 62, 83 and y where y is the missing value. According to the formula:
[tex]65.4 = \frac{57+55+62+83+y}{5}[/tex]
[tex]65.4 = \frac{257+y}{5}\\ 257+y = 65.4*5\\257+y = 327\\y = 327-257\\y = 70[/tex]
The missing value is 70
b) Since the total list of numbers is n values with a specific known mean, if some of this values can be freely selected, the number of n values that can be freely assigned before the remaining values are determined is any values less than n i.e n-1 values.
From the formula for calculating mean:
[tex]\overline x = \frac{\sum Xi}{N}\\{\sum Xi} = N\overline x[/tex]
This shows that the sum of all the values is equal to the product of the total values and the mean value. Since we can freely choose n-1 values, then sum of the set of data can be written as [tex]\sum \sumx^{n-1} _i__=_1 Xi[/tex]
[tex]\sum \sumx^{n-1} _i__=_1 Xi \ =\ N \overline x[/tex]
The number of n values which is referred to the degree of freedom is n-1
A cooler has a temperature of 32 degrees Fahrenheit. A bottled drink is placed in the cooler with an initial temperature of
70 degrees Fahrenheit. The function
[tex]f(t) = {ce}^{ (- kt)} + 32[/tex]
,represents the situation, where t is time in minutes, C is a
constant, and k is a constant.
After 3 minutes the bottle has a temperature of 42 degrees. What is the approximate value of k?
Answer:
[tex]k \approx 0.44[/tex]
Step-by-step explanation:
Given function:
[tex]f(t) = (ce)^{-kt}+32[/tex]
As per question statement:
Initial temperature of bottle is 70 [tex]^\circ F[/tex].
i.e. when time = 0 minutes, f(t) = 70 [tex]^\circ F[/tex]
[tex]70 = ce^{-k\times 0}+32\\\Rightarrow 38 = ce^{0}\\\Rightarrow c = 38[/tex]
After t = 3, f(t) = 42[tex]^\circ F[/tex]
[tex]42 = 38 \times e^{-k\times 3}+32\\\Rightarrow 42-32 = 38 \times e^{-3k} \\\Rightarrow 10 = 38 \times e^{-3k} \\\Rightarrow e^{3k} = \dfrac{38}{10}\\\Rightarrow e^{3k} = 3.8\\\\\text{Taking } log_e \text{both the sides:}\\\\\Rightarrow log_e {e^3k} = log_e {3.8}\\\Rightarrow 3k \times log_ee=log_e {3.8} (\because log_pq^r=r \times log_pq)\\\Rightarrow 3k \times 1=log_e {3.8}\\\Rightarrow 3k = 1.34\\\Rightarrow k = \dfrac{1.34}{3}\\\Rightarrow k \approx 0.44[/tex]
Hence, the value is:
[tex]k \approx 0.44[/tex]
P(AB) can be read as "the probability that A occurs given that Bhas
occurred."
A. True
B. False
Answer:
False
Step-by-step explanation:
from *millermoldwarp*
"Events are called dependent when the probability of an event depends on the occurrence of another. When event A depends on event B, the probability that A occurs, given that B has occurred, is different from the probability that A occurs only ."
hopes this helps
Answer:false
Step-by-step explanation: